Let $\gamma: [0,1] \to \{z \in \mathbb C: Im(z) \ge 0 \}$ be a closed and rectifiable path such that $\gamma(0) = 1, \gamma(1)=-1$ and ..

Question

Let $\gamma: [0,1] \to \{z \in \mathbb C: Im(z) \ge 0 \}$ be a closed and rectifiable path such that $\gamma(0) = 1, \gamma(1)=-1$ and $\gamma(t) \neq 0$ for all $t \in [0,1]$. Suppose $f$ is a holomorphic function on an open ball centered at 0 that contains $\{\gamma \}$ and such that $f(e^{it}) = e^{2it} + 1, \pi \le t \le 2\pi$ and $f'(0)=0$. Calculate $\int_{\gamma} \frac{f(z)}{z^2}dz.$

How do I solve this integral? I'm having a hard time even getting started. Does anyone know of a complex analysis book with examples of similar exercises. I'm using Conway and it doesn't have many examples.

Answer 1

It's not a full answer, but it's too long for a comment.

Let $B_r(0)$ be "the open ball centered at $0$, that $f$ is holomorphic on. Since it contains $\{\gamma\}$ and $\gamma(0) =1$, we get $r>1$. Further, let $A=\{e^{it}\mid t\in [\pi, 2\pi]\}$, then $f(z)=z^2+1$ for all $z\in A$ by definition. Observe, that $|z|=1$ for all $z\in A$, which means $A\subset B_r(0) $. Moreover, all points of $A$ are accumulation points, which leads to $f(z)=z^2+1$ for all $z\in B_r(0)$ by the identity theorem https://en.wikipedia.org/wiki/Identity_theorem

Then, \begin{align} \int_{\gamma}\frac{f(z) }{z^2}dz=\int_{\gamma}1+\frac{1}{z^2}dz=\left.z-\frac 1z\right|_1^{-1}=0 \end{align}

Hopefully, there is no mistake in it.

Answer 2

Since $\gamma(t)\neq 0$, it's contained in a (closed) upper half disk minus a small ball at the origin, which we can assume is contained in the domain of $f$. Call this set $E$. Since $f$ is holomorphic in $E$, and $E$ is simply connected, the integral over $\gamma$ is path-independent. In particular we can take $\gamma(t) = e^{-i\pi t}$ and evaluate the integral directly (noting that $f(e^{-i\pi t}) = e^{-2i\pi t}+1$ for $t\in [0,1]$):

$$ \int_{\gamma} \frac{f(z)}{z^2} dz = \int_0^1 \frac{e^{-2i\pi t}+1}{e^{-2i\pi t}} (-i\pi e^{-i\pi t})dt = -i\pi \int_0^1 (e^{-i\pi t}+e^{i\pi t}) dt = -2\pi i \int_0^1 \cos(\pi t)dt = -2 i [\sin(\pi)-\sin(0)] = 0$$